1969 AHSME Problems/Problem 19: Difference between revisions
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== Solution == | == Solution == | ||
<math>\fbox{B}</math> | Let <math>(xy)^2=a</math>. The expression given is equal to <math>a^2-10a+9=0</math>, which can be factored as <math>(a-9)(a-1)=0</math>. Thus, we have <math>a=(xy)^2=9</math> and <math>a=(xy)^2=1</math>. Because <math>x</math> and <math>y</math> are positive, we can eliminate the possibilities where <math>xy</math> is negative. From here it is easy to see that the only integral pairs of <math>x</math> and <math>y</math> are <math>(3, 1), (1, 3)</math>, and <math>(1, 1)</math>. The answer is <math>\fbox{B}</math>. | ||
== See also == | == See also == | ||
Latest revision as of 15:14, 10 July 2015
Problem
The number of distinct ordered pairs 
where 
and 
have positive integral values satisfying the equation 
is:
Solution
Let 
. The expression given is equal to 
, which can be factored as 
. Thus, we have 
and 
. Because and are positive, we can eliminate the possibilities where 
is negative. From here it is easy to see that the only integral pairs of and are 
, and 
. The answer is 
.
See also
| 1969 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
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